Mix/max of sin^3(x) - cos^2(x)

[bcahmel]

Homework Statement

Find the min and max of sin^3(x) - cos^2(x) on the interval [0,2pi]

The Attempt at a Solution

So I took the derivative, which is 3sin^2(x)cos(x) + 2cos(x)sin(x)
Then I set it to 0 and factored to get the crit pts:
0=(cosxsinx)(3sinx+2)

so cosx=0, sinx=0, and 3sinx+2=0
so x=pi, 2pi, 0, and -0.6184

And plugging these values into get the y's I got: -1,-1,-1,-0.8587


So is the max -1 and mix -0.8587? I think I messed up somewhere. Any help?

 

[micromass]

For what values is cos(x)=0 (there are 2 values of x in [0,2pi])
For what values is sin(x)=0 (there are 3 values of x in [0,2pi])
For what values is sin(x)=-2/3 (there are 2 values of x in [0,2pi])

 

[emailanmol]

Find where cosx =0,sinx=0 and sinx=-2/3 as mentioned above

Then put these in equation sin^3(x) - cos^2x as you want the max,min of the equation not cos and sin

Chose the two appropriate values

Question Solved

Anmol

 

[bcahmel]

So...
cosx=0 at pi/2, 3pi/2
sinx=0 at 0, pi, 2pi
sinx=-2/3 at -0.7297, ahh I forget trig,how do you get the second value?

 

[micromass]

Think of the following formula:

\sin(\pi-\alpha)=\sin(\alpha)

 

[bcahmel]

so sin(pi+0.7297) =3.8712??

 

[bcahmel]

I guess for the min I got -1 and the max I had 1. Is this right?

 

[micromass]

Yes, so your two solutions of \sin(x)=-3/2 are correct:

x=3.8712~\text{and}~x=-0.7297

I have only one stupid remark. The last value is not in [0,2pi], and we do want values in that interval...

 

[micromass]

I guess for the min I got -1 and the max I had 1. Is this right?

This is correct!

 

[bcahmel]

ok, thank you so much! You really helped! The answer should have been obvious I guess- since sin and cos both have max and min of 1 and -1 anyway for their range.